Difference between revisions of "Aufgaben:Problem 8"

Revision as of 15:16, 2 January 2015

Concerning part a)

First thing: It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \).

Wouldn't it be much easier to just:

$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$

and then use Bernoulli-L'Hopital to get:

$$ \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$

Second thing: Nevermind, confused myself...

Greetings A.

I didn't write that: I meant that the limit of the different quotient is the derivative. I was a bit inconsistent with the variable x (now it's fixed, I've used a y). If you prefer Bernoulli it's the same for me. Nick

Okay, I see what you meant now. I personally find it a bit confusing. But I guess it's a matter of taste.

A.

To Part b)

Question:

Since we had an whole exercise sheet on Lebesgue-convergence and i would assume they would ask of us to find an upper bound function g to prove that we're allowed to use that theorem. Why don't you do that? (It might be obvious but those just learning it by heart won't notice.)

Answer:

The upper bound function is for dominated convergence, not for monotone convergence. But I agree it is not so clearly stated, I'll fix it, thanks